Figure 1.3: Vector components for electric fields from two charges.
| (1.2.0.6) |
That is to say, the force on a charge q which is in the presence of an electric field E is F = qE. Note that the bold letters used indicate that E and F are vectors. The relationship we have used makes it easy to define the properties of E in terms of the point charges that produce them. Since we have already seen how one point charge (the charge, say on an elementary particle) can exert a force on a second (see the previous lecture), we should state that the electric field produced by a positive (negative) point charge goes radially outward (inward) in all directions from (toward) the charge.
The program shows the correct convention for drawing electric field lines. The field lines are a (rather gross) characterization of the actual electric field. The field extends radially, in all directions in 3-dimensional space, from the charge. If the charge is positive, the field is said to point away from the charge. If the charge is negative, the field points in towards the charge. The field lines give a general idea of the direction, but it is important to remember that the actual field in 3 dimensions extends to every point in space out to infinity if there are no other charges or matter.
The weird shape of the field for two or more charges comes about strictly for geometrical reasons. The electric force, we find experimentally, is both conservative and obeys the principal of superposition (just as gravitational forces do). Therefore, electric fields must also obey these rules. The superposition property means that electric fields from two different sources add vectorially to produce a net field. Let's look at two point charges as in lecture 1. The four possible combinations of positive or negative charges are shown.
Note that we have chosen the same position (relative to the charges) to calculate the net electric field in each of the four cases. For cases 1 and 2 (both charges the same sign), we note that the horizontal or x components of the fields tend to cancel, leaving the y components to add. This will be true for any point which is close to the line which is parallel to y and located at an x position close to the midpoint between the charges. If you look at the applet or in the textbook, you should note that the field lines indicate that the field is nearly vertical for points close to the midpoint line as we have described it. For cases 3 and 4 (unlike charges), the position we have chosen for calculating the net electric field tends to have the y components canceling and the x components adding. Hence any point near the midpoint line for unlike charges will have electric fields pointing nearly horizontally. Right along the midpoint line, the y component is exactly zero (assuming the two charges are equal magnitude). Again, you should verify that the applet and the textbook images impart the same "idea" to you as to what the field direction is near the midpoint line.
To draw the field lines for two charges as in the text or in the Java applet, we need to find the vector sum of the electric fields extending radially from or toward each charge (depending on whether the charge is positive or negative) at each point in space, then, for some small subset of points, find the direction of the net electric field. The field lines in the book or in the applet show the general direction of the electric field for various positions around the charges. Thus, the shape looks complex, but it is arrived at by doing the vector sum of radial field lines from each charge, so nothing complex is going on.
Assume that the red particles are positively charged and the blue particle is negatively charged. Since the particles are along the vertices of an equilateral triangle, the relative angles with respect to a convenient x-y axis is easy to determine. We observe the following for the contributions to the electric field at the lower left corner:
Note that only the fields due to q2
and q3 have to be considered
as the force on q1 is determined
only by external charges (for the case of point particles).
Hence we can find the components of the net electric field at the
position of q1 as follows:
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As an example of the above statement, consider a charged line
segment of length l and charge +Q. Here, we have not a point,
but an infinitely thin line of charge. Without any other
information to go on, we assume that +Q is uniformly distributed
throughout the line. If we wish to know what the electric field
is at a point a distance h above the midpoint of the line, how
do we calculate this?
Figure 1.6: A linearly distributed charge.
We approach this problem by breaking the line into small segments of infinitesimal length. The lengths are small enough that we can assume that each little segment can be considered as a point (later we can make a self-consistent check of this assumption). Before we go through any calculation though, we consider how symmetry can reduce the amount of work we need to do. In this case, if we consider the result of two infinitesimal segments, one on either side of the midpoint line as shown below, we note that, as in the previous case of two point charges, the horizontal components of the electric fields tend to cancel in making the net field.
Thus, we note that we only need to keep track of the vertical components for the line segments as the horizontal ones will cancel out. Furthermore, we can get away with only adding up the contributions to the net electric field from one half of the line segment and multiplying by two to take care of the contribution of the other half. This is not much work in this case, but it's nice to be clever when you're sure you're right. In this case, the integration can proceed once we know how to write down dE. To get that, we need to consider the charge of the infinitesimal line segment. If we assume that the charge is uniformly distributed, then the charge of each infinitesimal segment must be
| (1.2.3.9) |
Another way of saying this is to consider the linear charge density as being Q/l. Then the charge of any length, dx, of the line segment is (Q/l) dx. The vertical component of the electric field due to any little length, dx, is therefore
| (1.2.3.10) |
where r2 = (x2 + h2). Now we can set up the integration and finish the problem. For convenience, let's place the origin of our coordinate system at the midpoint of the line segment. Maple makes quick work of the integral.
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Let's go back and check that we have self-consistency with our
original assumption about very short line segments looking like
point charges. If we consider the above answer, we can ask the
question: what does the electric field look like if l, the length
of the line segment, is itself very small compared to h. In other
words, if h is big, then the point where we want to calculate
the field is very far away from the segment and we expect that it
will be difficult to tell the difference between the field due to
a point charge and the field due to a small line segment. On the
other hand, if h is not very large but l is very small, again, we
assume the field looks like that due to a point charge. We can
check this answer by looking at the above answer for E and letting
h be much greater than l. For that case, the square root in the
denominator looks like
| (1.2.3.12) |
Send comments to larryg@upenn5.hep.upenn.edu.
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